What is the return value of f(p, p) if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value.
int f(int &x, int c) {
c = c - 1;
if (c == 0) return 1;
x = x + 1;
return f(x, c) * x;
}
(A) 3024
(B) 6561
(C) 55440
(D) 161051
Answer: (B)
Explanation: Since c is passed by value and x is passed by reference, all functions will have same copy of x, but different copies of c.
f(5, 5) = f(x, 4)*x = f(x, 3)*x*x = f(x, 2)*x*x*x = f(x, 1)*x*x*x*x = 1*x*x*x*x = x^4
Since x is incremented in every function call, it becomes 9 after f(x, 2) call. So the value of expression x^4 becomes 9^4 which is 6561.
#include <stdio.h>
int f( int &x, int c)
{
c = c - 1;
if (c == 0) return 1;
x = x + 1;
return f(x, c) * x;
}
int main()
{
int p = 5;
printf ( "%d" , f(p, p));
}
|
Quiz of this Question
Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape,
GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out -
check it out now!